[LeetCode] Next Greater Element II

Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. The Next Greater Number of a number x is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn't exist, output -1 for this number.

Example 1:

Input: [1,2,1]
Output: [2,-1,2]
Explanation: The first 1's next greater number is 2; 
The number 2 can't find next greater number; 
The second 1's next greater number needs to search circularly, which is also 2.

Note: The length of given array won't exceed 10000.

找出一个元素后面大于其值的第一个元素。数组是一个循环数组。

将每个元素的索引放入stack中进行比较，并且因为是循环数组，所以要循环两次该数组。

class Solution {
public:
    vector<int> nextGreaterElements(vector<int>& nums) {
        if (nums.empty())
            return {};
        int n = nums.size();
        vector<int> res(n, -1);
        stack<int> stk;
        for (int i = 0; i < 2 * n; ++i)
        {
            int num = nums[i % n];
            while (!stk.empty() && nums[stk.top()] < num)
            {
                res[stk.top()] = num;
                stk.pop();
            }
            if (i < n)
                stk.push(i);
        }
        return res;
    }
};